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修改 0704 二分查找 JavaScript版本解法

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原解法逻辑不够简洁清晰，简单问题被其复杂化，所以我提出个人认为更简洁清晰的解法
This commit is contained in:
Jerry-306
2021-09-24 15:59:16 +08:00
committed by GitHub
parent 0f0bb6e741
commit 431b5a7e02
1 changed files with 32 additions and 17 deletions
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49
problems/0704.二分查找.md
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@ -283,31 +283,46 @@ func search(nums []int, target int) int {
// （版本一）左闭右闭区间
/**
* @param {number[]} nums
* @param {number} target
* @return {number}
*/
var search = function(nums, target) {
let l = 0, r = nums.length - 1;
// 区间 [l, r]
while(l <= r) {
let mid = (l + r) >> 1;
if(nums[mid] === target) return mid;
let isSmall = nums[mid] < target;
l = isSmall ? mid + 1 : l;
r = isSmall ? r : mid - 1;
let left = 0, right = nums.length;
// 使用左闭右开区间 [left, right)
while (left < right) {
let mid = left + Math.floor((right - left)/2);
if (nums[mid] > target) {
right = mid; // 去左区间寻找
} else if (nums[mid] < target) {
left = mid + 1; // 去右区间寻找
} else {
return mid;
}
}
return -1;
};
// （版本二）左闭右开区间
/**
* @param {number[]} nums
* @param {number} target
* @return {number}
*/
var search = function(nums, target) {
let l = 0, r = nums.length;
// 区间 [l, r）
while(l < r) {
let mid = (l + r) >> 1;
if(nums[mid] === target) return mid;
let isSmall = nums[mid] < target;
l = isSmall ? mid + 1 : l;
// 所以 mid 不会被取到
r = isSmall ? r : mid;
let left = 0, right = nums.length;
// 使用左闭右开区间 [left, right)
while (left < right) {
let mid = left + Math.floor((right - left)/2);
if (nums[mid] > target) {
right = mid; // 去左区间寻找
} else if (nums[mid] < target) {
left = mid + 1; // 去右区间寻找
} else {
return mid;
}
}
return -1;
};